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I need a clarification of one of the consequences of archimede's principle. My main doubt is why is it that the pressure on a certain object is independent of that object? This makes total sense, just thinking in various real life scenarios, howevever the reason for it isn't clear to me. Let's say we submerge an object on a cylinder of water, with volume v and we have another cylinder of water with the same volume. In a certain height h the pressure is the same, this pressure comes from weight of the mass of water above divided by the area. But aren't the weights of the mass water above h different? My view is the following: In the case 1, the buyoant force made the fluid go up, however because the mass of water didn't change, same number of atoms, it only means the atoms are more dispersed, while in the other case we've added mass. So how come the pressure in a certain dept h is the same in both cases, if we have different masses originating the force of the pressure?

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I may have misunderstood your explanation but I have the impression that what is wrong with your reasoning is that you are confusing pressure at a height $h$ and the buoyancy, which involves the variation in pressure over the entire submerged solid.

In the case of a vertical cylinder, things are simple. To find the buoyancy, we must take into account the pressure on the top of the cylinder and that on the base of this cylinder. These two pressures can vary with the depth $h$ but the pressure difference between the top and the bottom is not changed (if the fluid is incompressible) : so the buoyancy does not depend on $h$.

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Archimede's principle comes as a consequence of Stevino's law for fluid statics, and the laws

Let's focus here for simplicity on fluids with constant density $\rho$ and constant volume force $\mathbf{g}(\mathbf{r}) = g \mathbf{\hat{z}}$, representing your set-up of a vessel filled with water close to Earth surface with $z$-axis pointing downwards.

Under these assumptions, Stevino's law reads

$P(\mathbf{r}) = P_0 + \rho g z$.

The overall force acting on the immersed body, occupying volume $V$, due to the fluid, comes from the stress integrated over the whole surface $\partial V$ of the body

$\displaystyle \mathbf{F} = \oint_{\partial V} \mathbf{t_n} = - \oint_{\partial V} P \mathbf{\hat{n}} = - P_0 \oint_{\partial V} \mathbf{\hat{n}} - \rho g \oint_{\partial V} z \, \mathbf{\hat{n}}$,

and using the gradient theorem, $\displaystyle \oint_{\partial V} f \mathbf{\hat{n}} = \int_V \nabla f$,

$\displaystyle \mathbf{F} = -P_0 \int_V \nabla 1 - \rho g \int_V \nabla z = 0-\rho g \int_V \mathbf{\hat{z}} = -\rho g V \mathbf{\hat{z}}$,

i.e. equals the weight of the volume of the displaced fluid.

Note. Remember that we chose the $z$-axis pointing downwards, so the results we get agrees with the experience, with the buoyant force pointing upwards.

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    $\begingroup$ Bodies with the same shape and dimensions move the same volume of fluid. The fluid surrounding the body knows nothing about the body itself, beside its volume $\endgroup$
    – basics
    Dec 11, 2022 at 14:47
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    $\begingroup$ Force doesn't come mass of water above the body, but on the pressure acting on the surface of the body. At a larger depth, the body has more mass of water above and thus larger pressure on the upper surface of the body pushing it downwards, but even the lower surface of the body feels a higher pressure, pushing the body upwards. The overall contribution of stresses gives you Archimedes' law $\endgroup$
    – basics
    Dec 11, 2022 at 14:55
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    $\begingroup$ pressure acting on the body at different depths differs by a constant term given by $\rho g \Delta h$, being $\Delta h$ the different depths of the same reference point. This constant term, as the constant contribution due to $P_0$ vanishes when you perform integration over the whole surface. $\endgroup$
    – basics
    Dec 11, 2022 at 14:57
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    $\begingroup$ If you have a rigid body, you don't care about the depth, since the buoyancy is independent on the depth, if the body is fully submerged. When you have a deformable body, or you do underwater diving, you care about the overall pressure but "only" for physical limits of your body contrasting a high external pressure $\endgroup$
    – basics
    Dec 11, 2022 at 15:01
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    $\begingroup$ you're welcome. $\endgroup$
    – basics
    Dec 11, 2022 at 15:02
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So how come the pressure in a certain dept h is the same in both cases, if we have different masses originating the force of the pressure?

Does the buoyant force originate from the the floating mass? Or does it come from the water surrounding it?

If it originates from the water, shouldn't it depend on the geometry of the water surrounding the object?

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  • $\begingroup$ it does come from the water, but my point is the mass of the water is different in those cases, right? So how come can we have equal pressures being applied $\endgroup$ Dec 11, 2022 at 14:34

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